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Big Rip red sky?

  1. Nov 18, 2006 #1
    I'm curious about one thing. Suppose the universe will end in a Big Rip. Then since the universe will be accelerating at a very fast rate, could it be that the sky will look redder and redder towards the Big Rip?
  2. jcsd
  3. Nov 18, 2006 #2


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    The big-rip is a very strange thing. The scale factor (that describes the relation between the distance between to different points at different cosmological times) becomes infinite in a finite time. For this to take place, the energy density becomes infinite. This leads to an infinitely strong pressure which in turn leads to this infinite acceleration. This energy density is however the energy density of the phantom energy whereas the energy density of every other component in the universe gets diluted. For example, the cosmological photon background (microwave background) would get diluted and there is no way for it to increase its energy density.

    However, the big-rip is not the end of the universe. Time goes on after the big-rip and, in the usual big-rip scenarios, it starts a phase of contraction that leads to a big-crunch. This assumes that we can extrapolate the classica physics over the infinite density state of the big-rip that cannot be described currently. During this contraction phase the energy density of all other components of the universe increases and the photon background would get more and more energetic.
    Last edited: Nov 19, 2006
  4. Nov 18, 2006 #3
    Okay, then we know some more about the big rip. Just one question, what would cause this infinit energy? I suppose (from what I read) that you mean dark energy. That the cosmological constant describes a force that makes objects move away from eachother in increasing speed, and that this constant would go bazoonka. Or am I totally wrong?
  5. Nov 19, 2006 #4


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    A cosmological constant has constant energy density in time with an equation of state [itex]p = - c^2 \rho[/itex]. For the big-rip to take place a variable energy density is required with an equation of state [itex]p < - c^2 \rho[/itex]. In such a case one speaks about "phantom energy". For the usual big-rip scenarios, this energy density increases becoming infinite in finite time, leading to a state that cannot be described with current physics.
  6. Nov 19, 2006 #5
    Okay, moving on. The thought is that we are attracted by a higher energy density along a fourth coordinate and therefor move towards it? This makes little sence in my oppinion since if a photon has no time vektor, how can it stay in our universe? And how is the energy emitted if the vektors along the fourth dimension does not vary in sign? Cause clearly, time passes and it doesn't in a photon which means that when the photon was created the time vektor became zero and it can only become that if there was a collision, and the time vektor was transformed into a room vektor. Makes sence? Time is bidirectional, and how big is the risk that we meet higher energy density in both direction. Clearly time is infinitely thin since we move along it in the speed of light. Only in this way can any amount of mass become energy, and only in this way time can be experienced as a non-roomly vektor. The vektor sum should always be preserved. And if the particles moving in one direction of time, bends towards a high energy density, then the particles in the other direction would not experience the same vektor as infinitely thin. They would leave eachothers universe.

    This is my oppinion, and it is my oppinion that I am aloud to express a such.
  7. Nov 19, 2006 #6
    I don't know that big rips require INFINITE expansion in finite time. Can't it just be a continual increase in the expansion rate that eventually begins to rip apart galaxies, solar systems, planets, molecules, and finally nuclei. If it were a constant dark energy density causing the universe to accelerate its expansion, then wouldn't the ripping stop where the binding energy (gravitation or otherwise) equals the dark energy density?

    Oh, and thanks for the support in other threads. :cool:
  8. Nov 19, 2006 #7


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    To me it seams that it is necessary that the scale factor becomes infinite in finite time, at least for a flat universe. I am not sure what happens in an universe which is not spatially flat.

    It follows from this argumentation:

    First, let's show that the phantom energy implies a growing Hubble parameter in a flat universe. The time derivative of the Hubble parameter is:

    [tex]\dot H = \frac{d}{dt} \left( \frac{\dot a}{a} \right) = \frac{\ddot a}{a} - \left( \frac{\dot a}{a} \right)^2[/tex]

    Let's take the Friedmann equations for zero spatial curvature [itex]k = 0[/itex] and insert in the relation above:

    [tex]\dot H = - 4 \pi G \left( \rho + \frac{p}{c^2} \right)[/tex]

    If we have a phantom energy:

    [tex]\frac{p}{c^2} < - \rho[/tex]


    [tex]\dot H > 0[/tex]

    Second, let's go back to the first relation for [itex]\dot H[/itex]. Take:

    [tex]a = t^q[/tex]

    Inserting for [itex]a[/itex], [itex]\dot a[/itex] and [itex]\ddot a[/itex] and imposing the condition [itex]\dot H > 0[/itex], we get:

    [tex]q (q-1) - q^2 > 0[/tex]

    This can only hold if [itex]q < 0[/itex], which in turn implies some asymptotic behaviour of [itex]a[/itex].

    By the way, the relation:

    [tex]a = t^q[/tex]

    with negative [itex]q[/itex] does not make sense at all, because [itex]a[/itex] must be zero at [itex]t = 0[/itex], but one could imagine a similar relation like e.g.

    [tex]a = - a_0 + (t - t_{R})^q[/tex]

    which needs of a negative [itex]q[/itex] to fulfil the phantom energy condition and has an asymptotic behaviour.

    From such a relation you can see also that there exists another branch for the scale factor after tR with a contraction. It is not clear, however, how to interpret this, because the universe goes through a singularity in the energy density (from the first Friedmann equation you can see how the energy density behaves).
    Last edited: Nov 19, 2006
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